The curve C has equation $y=\dfrac{2x^{2}-a}{x+k}$ where $a,\; k>0$.

1. Find the derivative of $y$ and set it to $0$ to locate points where the tangent is parallel to the $x$-axis.

Using the quotient rule:
$\frac{dy}{dx} = \frac{(x+k)(4x) - (2x^2-a)(1)}{(x+k)^2}$
$\frac{dy}{dx} = \frac{4x^2 + 4kx - 2x^2 + a}{(x+k)^2} = \frac{2x^2 + 4kx + a}{(x+k)^2}$
For the curve to have a tangent parallel to the $x$-axis, $\frac{dy}{dx} = 0$, which means the numerator must be zero:
$2x^2 + 4kx + a = 0$

2. Apply the discriminant condition for the quadratic equation to yield real roots.

For the curve to have at least one tangent parallel to the $x$-axis, the quadratic equation $2x^2 + 4kx + a = 0$ must have real roots. Thus, its discriminant must be non-negative ($\Delta \ge 0$):
$(4k)^2 - 4(2)(a) \ge 0$
$16k^2 - 8a \ge 0$
$2k^2 - a \ge 0$

3. Solve the inequality for $k$, applying the given conditions $2k^2 \ne a$ and $k > 0$.

Since it is given that $2k^2 \ne a$, we must strictly have:
$2k^2 - a > 0 \implies k^2 > \frac{a}{2}$
(Note: Checking if the stationary points can land on the vertical asymptote $x = -k$: substituting $x = -k$ gives $2(-k)^2 + 4k(-k) + a = a - 2k^2 \ne 0$. Thus, the points are valid.)
Given that $a > 0$ and $k > 0$:
$k > \sqrt{\frac{a}{2}}$

1. Substitute $a = 2k^2$ into the equation and simplify.

$y = \frac{2x^2 - 2k^2}{x+k}$
$y = \frac{2(x^2 - k^2)}{x+k}$
$y = \frac{2(x-k)(x+k)}{x+k}$
$y = 2(x-k) \text{ for } x \neq -k$

2. Describe the resulting graph geometrically, accounting for the restricted domain.

Since the original function is undefined at $x = -k$, the graph is a straight line with the equation $y = 2x - 2k$ but with a hole at $x = -k$.
When $x = -k$, $y = 2(-k - k) = -4k$.
Thus, the graph is a straight line $y = 2x - 2k$ excluding the point $(-k, -4k)$.

1. Determine the equations of the vertical and oblique asymptotes.

By algebraic long division, rewrite the equation of the curve $C$:
$y = \frac{2x^2-a}{x+k} = \frac{2x(x+k) - 2k(x+k) + 2k^2 - a}{x+k} = 2x - 2k + \frac{2k^2-a}{x+k}$
Equating the denominator to zero gives the vertical asymptote:
$x = -k$
As $x \to \pm\infty$, $\frac{2k^2-a}{x+k} \to 0$. Thus, the oblique asymptote is:
$y = 2x - 2k$

2. Find the exact coordinates of the intercepts with the axes.

For $x$-intercepts, set $y = 0$:
$2x^2 - a = 0 \implies x^2 = \frac{a}{2} \implies x = \pm\sqrt{\frac{a}{2}}$
The $x$-intercepts are $\left(-\sqrt{\frac{a}{2}}, 0\right)$ and $\left(\sqrt{\frac{a}{2}}, 0\right)$.

For the $y$-intercept, set $x = 0$:
$y = \frac{-a}{k} = -\frac{a}{k}$
The $y$-intercept is $\left(0, -\frac{a}{k}\right)$.

3. Deduce the shape and layout of the curve to complete the sketch.

From part (i), the required range is $2k^2 > a$. Because $a > 0$ and $k > 0$, this means $k > \sqrt{\frac{a}{2}}$.
Since $-k < -\sqrt{\frac{a}{2}}$, both $x$-intercepts lie to the right of the vertical asymptote $x = -k$.
Because $2k^2 - a > 0$, the remainder in $y = 2x - 2k + \frac{2k^2-a}{x+k}$ is positive.
This means the curve approaches the oblique asymptote from above as $x \to \infty$, and from below as $x \to -\infty$.
Thus, the graph consists of a top-right branch (passing through all intercepts) and a bottom-left branch.

1. Identify the asymptotes and positional regions of both curves.

From part (i), Curve C is $y = 2x - 2k + \frac{2k^2-a}{x+k}$. Its oblique asymptote is $y = 2x - 2k$, which passes through $(-k, -4k)$ and has a gradient of $2$.
Since $2k^2 > a$, Curve C approaches and lies above this oblique asymptote for $x > -k$.

The given hyperbola is $(x+k)^2 - \left(\frac{y+4k}{b}\right)^2 = 1$, which is centred at $(-k, -4k)$ and opens horizontally.
Its asymptotes are given by $(x+k)^2 - \left(\frac{y+4k}{b}\right)^2 = 0 \implies y + 4k = \pm b(x + k)$, meaning the asymptote with a positive gradient has a gradient of $b$.
The right branch of the hyperbola (where $x > -k$) lies bounded vertically between these asymptotes, so it sits strictly below its upper asymptote $y + 4k = b(x + k)$.

2. Compare the asymptotes to determine the range of values of $b$ for non-intersection.

When drawn on the same diagram, the oblique asymptote of Curve C and the asymptotes of the hyperbola all intersect at the same point $(-k, -4k)$.
For the two curves not to intersect, the hyperbola must not overlap into the space occupied by Curve C.
For $x > -k$, Curve C is bounded below by a line of gradient $2$, while the hyperbola is bounded entirely above by a line of gradient $b$.
Therefore, the hyperbola's upper asymptote must not be steeper than the oblique asymptote of Curve C. Thus:
$b \le 2$
Given the condition that $b > 0$, the required range of values is:
$0 < b \le 2$